/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Total plasma volume is important... [FREE SOLUTION] | 91Ó°ÊÓ

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Total plasma volume is important in determining the required plasma component in blood replacement therapy for a person undergoing surgery. Plasma volume is influenced by the overall health and physical activity of an individual. (Reference: See Problem 16.) Suppose that a random sample of 45 male firefighters are tested and that they have a plasma volume sample mean of \(\bar{x}=37.5 \mathrm{ml} / \mathrm{kg}\) (milliliters plasma per kilogram body weight). Assume that \(\sigma=7.50 \mathrm{ml} / \mathrm{kg}\) for the distribution of blood plasma. (a) Find a \(99 \%\) confidence interval for the population mean blood plasma volume in male firefighters. What is the margin of error? (b) What conditions are necessary for your calculations? (c) Interpret your results in the context of this problem. (d) Sample Size Find the sample size necessary for a \(99 \%\) confidence level with maximal margin of error \(E=2.50\) for the mean plasma volume in male firefighters.

Short Answer

Expert verified
(a) The confidence interval is (34.62, 40.38) ml/kg, with a margin of error of 2.88 ml/kg. (b) A random sample and known standard deviation justify the calculation. (c) We are 99% confident the mean plasma volume is between 34.62 and 40.38 ml/kg. (d) A sample size of 75 is needed for a margin of error of 2.50.

Step by step solution

01

Identify the Problem Context

We need to find a confidence interval for the population mean plasma volume in male firefighters. We have a sample mean \( \bar{x} = 37.5 \mathrm{ml/kg} \), sample size \( n = 45 \), and known population standard deviation \( \sigma = 7.50 \mathrm{ml/kg} \).
02

Determine the Z-score for 99% Confidence Level

For a 99% confidence level, we need to determine the critical z-value. A 99% confidence level leaves 0.5% in each tail of the standard normal distribution. Using a standard normal table or calculator, we find that the z-score is approximately \( z = 2.576 \).
03

Calculate the Margin of Error

The margin of error (E) is calculated using the formula: \[ E = z \times \left( \frac{\sigma}{\sqrt{n}} \right) \]Substituting known values, we get:\[ E = 2.576 \times \left( \frac{7.50}{\sqrt{45}} \right) \approx 2.88 \]
04

Establish the Confidence Interval

The 99% confidence interval is given by:\[ \left( \bar{x} - E, \bar{x} + E \right) \]Substitute \( \bar{x} = 37.5 \) and \( E = 2.88 \):\[ (37.5 - 2.88, 37.5 + 2.88) = (34.62, 40.38) \]
05

Define Conditions for Valid Calculations

The necessary conditions are that the sample of male firefighters is random, the population standard deviation is known, and the sample size is large enough (n > 30) to justify using the normal distribution.
06

Interpret the Confidence Interval

We can say with 99% confidence that the true mean plasma volume for male firefighters is between 34.62 ml/kg and 40.38 ml/kg of body weight.
07

Compute Sample Size for Desired Margin of Error

To find the sample size (n) for a desired margin of error \( E = 2.50 \), use the formula: \[ n = \left( \frac{z \times \sigma}{E} \right)^2 \]Substitute \( z = 2.576 \), \( \sigma = 7.50 \), and \( E = 2.50 \): \[ n = \left( \frac{2.576 \times 7.50}{2.50} \right)^2 \approx 74.64 \]Since the sample size must be a whole number, round up to 75.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Population Mean
The concept of the population mean is central to understanding statistics. It represents the average of a specific characteristic for an entire population. In our context, it means the average plasma volume per kilogram of body weight among all male firefighters. Since it's often impossible to measure every individual in a large group, we use a sample, like our 45 firefighters, to estimate the population mean.
  • A sample mean is an estimate of the population mean.
  • The more representative and larger the sample, the more accurate our estimate tends to be.
Remember, the population mean ( ) itself is unknown. We use calculations and statistical methods to estimate it based on the data we have.
Standard Deviation
Standard deviation, symbolized by , is a measure of variance or spread in a set of data. It tells us how much individual data points tend to differ from the mean.
  • A small standard deviation implies that most of the data points are close to the mean.
  • A larger standard deviation suggests more widespread data.
In our exercise, the standard deviation of plasma volume among male firefighters is given as 7.50 ml/kg. This indicates that the individual plasma volumes tend to vary around this measure from the average of 37.5 ml/kg. Having a known population standard deviation is crucial when calculating precise confidence intervals for the population mean.
Margin of Error
The margin of error (E) represents the extent of uncertainty associated with a sample estimate of a population parameter. It helps define the confidence interval, which predicts a range within the population mean likely falls.
  • A larger margin of error suggests less precision in your estimate of the population mean.
  • The formula for margin of error in a confidence interval is: \[ E = z \times \left( \frac{\sigma}{\sqrt{n}} \right) \]
In our exercise, we calculated the margin of error for a 99% confidence interval as 2.88 ml/kg, given a z-score of 2.576 and sample size of 45. It's essential for interpreting how well our sample mean represents the true mean.
Sample Size Determination
Determining the right sample size is vital for obtaining reliable statistical results. The sample size affects the margin of error and, consequently, the confidence interval's accuracy. The formula for sample size determination to achieve a set margin of error is:\[ n = \left( \frac{z \times \sigma}{E} \right)^2 \]
  • Larger sample sizes generally lead to smaller margins of error, improving estimate precision.
  • For our desired margin of error of 2.50 ml/kg at a 99% confidence level, the calculation showed a needed sample size of 75.
A correctly determined sample size ensures the sample accurately reflects the population characteristics with specified levels of confidence and precision.

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Most popular questions from this chapter

In the Focus Problem at the beginning of this chapter, a study was described comparing the hatch ratios of wood duck nesting boxes. Group I nesting boxes were well separated from each other and well hidden by available brush. There were a total of 474 eggs in group I boxes, of which a field count showed about 270 had hatched. Group II nesting boxes were placed in highly visible locations and grouped closely together. There were a total of 805 eggs in group II boxes, of which a field count showed about 270 had hatched. (a) Find a point estimate \(\hat{p}_{1}\) for \(p_{1},\) the proportion of \(\mathrm{cggs}\) that hatched in group I nest box placements. Find a \(95 \%\) confidence interval for \(p_{1}\) (b) Find a point estimate \(\hat{p}_{2}\) for \(p_{2},\) the proportion of eggs that hatched in group II nest box placements. Find a \(95 \%\) confidence interval for \(p_{2}\) (c) Find a \(95 \%\) confidence interval for \(p_{1}-p_{2} .\) Does the interval indicate that the proportion of eggs hatched from group I nest boxes is higher than, lower than, or equal to the proportion of eggs hatched from group II nest boxes? (d) What conclusions about placement of nest boxes can be drawn? In the article discussed in the Focus Problem, additional concerns are raised about the higher cost of placing and maintaining group I nest box placements. Also at issue is the cost efficiency per successful wood duck hatch.

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